Answer :
Answer:
(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%
Explanation:
Using the formula that relate heat and work from the thermodynamic theory as:[tex]W=Q=Q_{in}-Q_{out}[/tex] solving to Q_out we get:[tex]Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ)[/tex] this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: [tex]n=1-\frac{T_{Low} }{T_{High}}[/tex] where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:[tex]n=1-\frac{423.15}{723.15} =0.415=41.5%[/tex]
Final answer:
A cyclical heat engine operating between temperatures of 450º C and 150º C dissipates 1.00 MJ of heat into the environment. The engine's efficiency is calculated to be 80%.
Explanation:
The question refers to the operation of a cyclical heat engine. Given the data, we can first calculate the amount of heat the engine dissipates into the environment using this formula: Qc = Qh - W, where Qh is the heat transfer into the engine (5.00 MJ) and W is the work done by the engine (4.00 MJ). When calculated, Qc = 5.00 MJ - 4.00 MJ = 1.00 MJ.
Next, let's calculate the engine's efficiency which can be defined as the ratio of the work output to the heat input, or e = W / Qh. Substituting the values from the problem, the efficiency = 4.00 MJ / 5.00 MJ = 0.80 or 80% when expressed in percentage terms.
So in answer to your questions, the heat transfer that occurs to the environment is 1.00 MJ and the engine's efficiency is 80%.
Learn more about Cyclical Heat Engine here:
https://brainly.com/question/30772526
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